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3 years ago

(-1) + 1 = 0 is which inverse

Mathematics

1 answer:

Licemer1 [7]3 years ago

7 0

Answer:

Heya mate....

Step-by-step explanation:

This is ur answer....

<h2>--> ADDITIVE INVERSE!</h2>

Additive inverse simply means changing the sign of the number and adding it to the original number to get an answer equal to 0. The properties of additive inverse are given below, based on negation of the original number. For example, x is the original number, then its additive inverse is -x.

Hope it helps you,

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3 years ago

A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 stude

Mkey [24]

Answer:

(a) Probability that 2 or fewer will withdraw is 0.2061.

(b) Probability that exactly 4 will withdraw is 0.2182.

(c) Probability that more than 3 will withdraw is 0.5886.

(d) The expected number of withdrawals is 4.

Step-by-step explanation:

We are given that a university found that 20% of its students withdraw without completing the introductory statistics course.

Assume that 20 students registered for the course.

The above situation can be represented through binomial distribution;

$P(X =r) = \binom{n}{r} \times p^{r} \times (1-p)^{n-r};x=0,1,2,3,......$

where, n = number of trials (samples) taken = 20 students

r = number of success

p = probability of success which in our question is probability

that students withdraw without completing the introductory

statistics course, i.e; p = 20%

Let X = Number of students withdraw without completing the introductory statistics course

So, X ~ Binom(n = 20 , p = 0.20)

(a) Probability that 2 or fewer will withdraw is given by = P(X $\leq$ 2)

P(X $\leq$ 2) = P(X = 0) + P(X = 1) + P(X = 2)

= $\binom{20}{0} \times 0.20^{0} \times (1-0.20)^{20-0}+ \binom{20}{1} \times 0.20^{1} \times (1-0.20)^{20-1}+ \binom{20}{2} \times 0.20^{2} \times (1-0.20)^{20-2}$

= $1 \times1 \times 0.80^{20}+ 20 \times 0.20^{1} \times 0.80^{19}+ 190\times 0.20^{2} \times 0.80^{18}$

= 0.2061

(b) Probability that exactly 4 will withdraw is given by = P(X = 4)

P(X = 4) = $\binom{20}{4} \times 0.20^{4} \times (1-0.20)^{20-4}$

= $4845\times 0.20^{4} \times 0.80^{16}$

= 0.2182

(c) Probability that more than 3 will withdraw is given by = P(X > 3)

P(X > 3) = 1 - P(X $\leq$ 3) = 1 - P(X = 0) - P(X = 1) - P(X = 2) - P(X = 3)

= $1-(\binom{20}{0} \times 0.20^{0} \times (1-0.20)^{20-0}+ \binom{20}{1} \times 0.20^{1} \times (1-0.20)^{20-1}+ \binom{20}{2} \times 0.20^{2} \times (1-0.20)^{20-2}+\binom{20}{3} \times 0.20^{3} \times (1-0.20)^{20-3})$